 Solution

Since $n_e=n_p=X_e{n}_b$ and $n_H=(1-X_e)n_b$,Using $n_b=\eta n_{\gamma }$ and $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ yieldsCosmological recombination occurs far below $B_H$ because the baryon-to-photon ratio is only about $10^{-9}$. There are roughly a billion photons per baryon, so the high-energy tail of the blackbody distribution continues to photoionize hydrogen until the exponential Boltzmann factor overcomes this enormous entropy factor.